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Chem 201 Final Exam: Filling in the Blanks and Balancing Redox Reactions - Prof. Timothy R, Exams of Chemistry

The final exam for a chemistry 201 course, consisting of multiple-choice questions and redox reaction balancing problems. Students are required to fill in the blanks with the correct answers and balance the given redox reactions. Topics covered include atomic orbitals, compounds' names, structures, and properties, as well as redox reactions and their balancing.

Typology: Exams

Pre 2010

Uploaded on 12/12/2009

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Chem 201 Name (PRINT!) KEY
Final Exam December 9, 2008
1. (23 pts) Fill in the blank with the best answer.
Hybrid atomic orbitals named sp2 are used by the central atom when it has 3
regions of electron density in the Lewis dot diagram.
An aldehyde would react with dichromate ion to yield the organic functional group called
Carboxylic acid (Be as specific as possible.)
A ketone would react with sodium borohydride to yield the organic functional group
Secondary alcohol (Be as specific as possible.)
63
Cu
2+
has 29 protons, 27 electrons, and 34 neutrons.
T, V, E, and H are examples of state functions, while q and w are
examples of path functions.
The quantum number n is called the primary quantum number; one of the
names of l is the secondary/azimuthal/subsidiary quantum number.
Oxides of metalloids are most likely to dissolve in solution described as
Acidic or basic (acidic/basic/either acidic or basic).
The energy released when a neutral gas phase atom gains an electron is called its
Electron affinity
Of the salts LiF, NaCl. and KBr, the one having the largest value for lattice energy would
be LiF .
2. (12 pts) Name the following compounds
Sr3(PO3)2 strontium phosphite
W(ClO3)3 tumgsten(III)chlorate
Hg2(C2H3O2) 2 mercurous acetate
H3AsO3 (aq) arsenous acid
/35
Chem 201 A Final Exam, p 2 Name
pf3
pf4
pf5
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Chem 201 Name (PRINT!) KEY

Final Exam December 9, 2008

  1. (23 pts) Fill in the blank with the best answer.

Hybrid atomic orbitals named sp2 are used by the central atom when it has 3

regions of electron density in the Lewis dot diagram.

An aldehyde would react with dichromate ion to yield the organic functional group called

Carboxylic acid (Be as specific as possible.)

A ketone would react with sodium borohydride to yield the organic functional group

Secondary alcohol (Be as specific as possible.)

63

Cu

2 + has 29 protons, 27 electrons, and 34 neutrons.

T, V, E, and H are examples of state functions, while q and w are

examples of path functions.

The quantum number n is called the primary quantum number; one of the

names of l is the secondary/azimuthal/subsidiary quantum number.

Oxides of metalloids are most likely to dissolve in solution described as

Acidic or basic (acidic/basic/either acidic or basic).

The energy released when a neutral gas phase atom gains an electron is called its

Electron affinity

Of the salts LiF, NaCl. and KBr, the one having the largest value for lattice energy would

be LiF.

  1. (12 pts) Name the following compounds

Sr 3 (PO 3 ) 2 strontium phosphite

W(ClO 3 ) 3 tumgsten(III)chlorate

Hg 2 (C 2 H 3 O 2 ) 2 mercurous acetate

H 3 AsO 3 (aq) arsenous acid

Chem 201 A Final Exam, p 2 Name

  1. (6 pts) Give an acceptable name for each of the following

O

OH

pentanoic acid

HO

2-butanol

NH 2

O

propanamide

  1. (4 pts) Draw the structure (similar to those in #3) for each of the following

methyl butanoate

O

O

cyclohexanone

O

  1. (10 pts) A photon with energy of 2.50 x 10

    Joules has 

a frequency of:

E

h

2.50x

  • J

6.626x

  • Js

= 3.77 x 10

16 s

a wavelength of:

λ =

c

ν

2.9979x

8 m s

3.77 x

16 s

= 7.95 x

  • m

Chem 201 A Final Exam, p 4 Name

  1. (27 pts) Calculate the values of N, A, and S for each of the following. Then write an

acceptable Lewis dot diagram, using o or x to represent every valence electron. If

resonance occurs, only one of the forms need be shown. Then indicate the electronic

geometry around the central atom, and the overall molecular geometry.

SF 2 N = 24 A = 20 S = 4

Hybrid orbitals for S = sp

Electronic geometry = tetrahedral

Molecular geometry = bent

TeF 5

N = 48 A = 42 S = 6

Hybrid orbitals for Te = sp3d

Electronic geometry = octahedral

Molecular geometry = square pyramidal

IF 4

N = A = S =

Hybrid orbitals for I = sp3d

Electronic geometry = trigonal bipyramidal

Molecular geometry = see-saw

Chem 201 A Final Exam, p 5 Name

  1. (18 pts) (a) Write the formation reaction of gaseous butane (i.e., the reaction whose

molar enthalpy change is termed

ΔHf

o )

4 C(s) + 5 H 2 (g) = C 4 H 10 (g)

(b) Write the balanced combustion reaction of one mole of butane gas with oxygen gas to

form carbon dioxide gas and liquid water.

C 4 H 10 (g) + 6.5 O 2 (g) = 4 CO 2 (g) + 5 H 2 O(l)

The molar enthalpy change of combustion of butane

ΔHcombustion

o is -2878 kJ/mole.

The

ΔHf

o of carbon dioxide gas is -393.5 kJ/mole, and the

ΔHf

o of liquid water

is -285.83 kJ/mole. Calculate the value of

ΔHf

o for butane gas.

ΔHreaction = Sum ΔH products – Sum of ΔH reactants

-2878 kJ/mole = [4(-393.5 kJ/mol) + 5(-285.83 kJ/mol)] – [ΔHf butane + 6.5 (0)]

ΔHf

o butane gas = +2878 - 3003 = - 125 kJ/mole

2 NaOH(s) + Cu

2+ (aq) = 2 Na

(aq) + Cu(OH) 2 (s)

Chem 201 A Final Exam, p 7 Name

  1. (25 pts) You must show all your work and prove any assumptions! Showing the

results in the form of a table is strongly encouraged. Assuming complete reaction, how

many moles remain at the end of the reaction for each reactant and each product when

40.00 g of phosphorous, 75.00 g of barium hydroxide, 20.00 g of water, and 15.00 g of

sulfuric acid are reacted according to the reaction:

2 P 4 + 3 Ba(OH) 2 + 6 H 2 O + H 2 SO 4

__

3 BaSO 4 + 6 H 3 PO 2 + 2 PH 3

40.00 g P4(1 mole/123.896g) = 0.3228 mol P

75.00 g Ba(OH)2 (1 mole/171.3416 g) = 0.4377 mole Ba(OH)

20.00 g H2O(1 mole H2O/18.01528 g) = 1.110 mole H2O

15.00 g H2SO4 (1 mol H2SO4/98.0794 g) = 0.1529 mole H2SO

2 P 4 + 3 Ba(OH) 2 + 6 H 2 O + H 2 SO 4

__

3 BaSO 4 + 6 H 3 PO 2 + 2 PH 3

mol inti .3228 .4377 1.110 .1529 0 0 0

mol react 2x 3x 6x x - - -

mol form - - - - 3x 6x 2x

left .3228- .4377- 1.110- .1529-

2x=0 3x=0 6x = 0 x = 0

x=? .1614 .1459 .1850.

L.R.

Left .0310 0 .235 .0070 .4377 .8754.

Chem 201 A Final Exam, p 8 Name

  1. (16 pts) The electron in a hydrogen atom moved from the n=3 to the n=5 orbit. The

photon associated with this transition is absorbed (absorbed/emitted).

Calculate the wavelength, in nm, of this photon.

1

λ

= R H

1

n f

2

1

n i

2

⎜ ⎜

⎟ ⎟

= 1.09737m

-1 1

5

2

1

3

2

⎜ ⎜

⎟ ⎟ = 7.803 x 10

5 m

λ =

1

7.803x

5 m

  • = 1.281x - m

10

9 nm

1 m

⎟ =^ 1281 nm

  1. (16 pts) A cup (heat capacity 15.0 J/K) and the sample of 25.0 g of water (specific

heat 4.18 J/g K) the cup contained were both initially at a temperature of 21.

o C.

A 50.0 g piece of lead, initially at a temperature of 175.

o C, was added to the water and

cup. The final common temperature was 28.

o C. What is the specific heat of lead?

0 = (15.0 J/K)(28.8-21.0deg) + (25.0g)(4.28J/gK)(28.8-21.0deg)

  • 50.0g(SpHt)(28.8-175.0deg)

0 = 117 J + 815 J - 7310 g K (SpHt)

SpHt = 0.127 J/gK